noi2017解题报告（部分）

D1T1:
每30位压成一个int，然后压位线段树就可以过了。

#include <bits/stdc++.h>#define gc getchar()#define N 4500009#define mid (l+r>>1)#define ll long longusing namespace std;const int Max=1<<30;int n,m,t1,t2,t3;struct node{    int val;    node *lc,*rc;    void up()    {        if (lc->val==rc->val) val=lc->val;        else val=-1;    }    void down()    {        if (val!=-1) lc->val=rc->val=val;    }    int ins(int l,int r,int x,int y,int ret=0)    {        if (l==r)        {            val+=y;            if (val>=Max)            {                val-=Max;                return 1;            }            if (val<0)            {                val+=Max;                return -1;            }            return 0;        }        if (val==0&&l==x&&y==-1)        {            val=Max-1;            return -1;        }        if (val==Max-1&&l==x&&y==1)        {            val=0;            return 1;        }        down();        if (x<=mid)        {            ret=lc->ins(l,mid,x,y);            if (ret) ret=rc->ins(mid+1,r,mid+1,ret);        }        else ret=rc->ins(mid+1,r,x,y);        up();        return ret;    }    int get(int l,int r,int x)    {        if (l==r) return val;        down();        if (x<=mid) return lc->get(l,mid,x);        else return rc->get(mid+1,r,x);    }}a[N],*now=a,*root;node *build(int l,int r){    node *Now=now++;    Now->val=0;    if (l==r) return Now;    Now->lc=build(l,mid);    Now->rc=build(mid+1,r);    return Now;}int read(){    int x=1;    char ch;    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;    int s=ch-'0';    while (ch=gc,ch<='9'&&ch>='0') s=s*10+ch-'0';    return s*x;}int main(){    m=read(),t1=read(),t2=read(),t3=read();    n=m+32;    root=build(0,n);    while (m--)    {        int op=read();        if (op==1)        {            int a=read(),b=read();            if (a>=0)            {                int a1=((ll)a<<(b%30))&(Max-1);                int a2=a>>(30-b%30);                int b1=b/30,b2=b1+1;                if (a1) root->ins(0,n,b1,a1);                if (a2) root->ins(0,n,b2,a2);            }            else            {                a=-a;                int a1=((ll)a<<(b%30))&(Max-1);                int a2=a>>(30-b%30);                int b1=b/30,b2=b1+1;                if (a1) root->ins(0,n,b1,-a1);                if (a2) root->ins(0,n,b2,-a2);            }        }        else        {            int a=read();            int ans=root->get(0,n,a/30);            printf("%d\n",(ans>>(a%30))&1);        }    }    return 0;}

D1T2：
根据miaom大的说法，因为拆开的操作很少，所以每次暴力维护所有len≤50的字符串就可以了，hash+hash表（注意双哈希）

#include <bits/stdc++.h>#define gc getchar()#define ll long long#define A 7#define MOD 5614657#define Mod 44917381#define mod 998244353#define N 200009#define M 10000009#define DEBUG 0#define debug() cerr<<"wrong"<<endlusing namespace std;int n,m,a[N],fa[N],son[N],b[M],p[55][2];struct HASH{    #define MM 10000010    int number,first[MOD];    struct edge    {        int y,next,val;        edge(int y=0,int next=0,int val=0):y(y),next(next),val(val){}    }e[MM];    inline void add(int x,int y)    {        e[++number]=edge(y,first[x],1);        first[x]=number;    }    inline void ins(int x,int y)    {        for (int i=first[x];i;i=e[i].next)            if (e[i].y==y)            {                e[i].val++;                return;            }        add(x,y);    }    inline void del(int x,int y)    {        for (int i=first[x];i;i=e[i].next)            if (e[i].y==y)            {                e[i].val--;                return;            }    }    inline int find(int x,int y)    {        for (int i=first[x];i;i=e[i].next)            if (e[i].y==y) return e[i].val;        return 0;    }}ha;inline int read(){    int x=1;    char ch;    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;    int s=ch-'0';    while (ch=gc,ch<='9'&&ch>='0') s=s*10+ch-'0';    return s*x;}inline void link(int x,int y){    fa[y]=x,son[x]=y;    int last=x,next=y,len1=1,len2=1,len=0;    for (int i=1;i<=48&&fa[last];i++) last=fa[last],len1++;    for (int i=1;i<=48&&son[next];i++) next=son[next],len2++;    for (int i=1;i<=len1+len2;i++)        b[++len]=a[last],last=son[last];    for (int i=1;i<=len1;i++)    {        int now=0,Now=0;        for (int j=i;j<=len1;j++)        {            now=(now*A%MOD+b[j])%MOD;            Now=(Now*A%Mod+b[j])%Mod;        }        for (int j=len1+1;j-i+1<=50&&j<=len;j++)        {            now=(now*A%MOD+b[j])%MOD;            Now=(Now*A%Mod+b[j])%Mod;            ha.ins(now,Now);        }    }}inline void cut(int x,int y){    int last=x,next=y,len1=1,len2=1,len=0;    for (int i=1;i<=48&&fa[last];i++) last=fa[last],len1++;    for (int i=1;i<=48&&son[next];i++) next=son[next],len2++;    for (int i=1;i<=len1+len2;i++)        b[++len]=a[last],last=son[last];    for (int i=1;i<=len1;i++)    {        int now=0,Now=0;        for (int j=i;j<=len1;j++)        {            now=(now*A%MOD+b[j])%MOD;            Now=(Now*A%Mod+b[j])%Mod;        }        for (int j=len1+1;j-i+1<=50&&j<=len;j++)        {            now=(now*A%MOD+b[j])%MOD;            Now=(Now*A%Mod+b[j])%Mod;            ha.del(now,Now);        }    }    fa[y]=son[x]=0;}inline int solve(){    int len=0,k,now=0,Now=0;    char ch;    while (ch=gc,ch<'1'||ch>'6');    while (ch!=' ') b[++len]=ch-'0',ch=gc;    k=read();    for (int i=1;i<=k;i++)    {        now=(now*A%MOD+b[i])%MOD;        Now=(Now*A%Mod+b[i])%Mod;    }    int ans=ha.find(now,Now);    for (int i=2;i<=len-k+1;i++)    {        now=(now-(ll)b[i-1]*p[k-1][0]%MOD+MOD)%MOD;        now=(now*A%MOD+b[i+k-1])%MOD;        Now=(Now-(ll)b[i-1]*p[k-1][1]%Mod+Mod)%Mod;        Now=(Now*A%Mod+b[i+k-1])%Mod;        ans=(ll)ans*ha.find(now,Now)%mod;        if (ans==0) break;    }    return ans;}int main(){    if (DEBUG)    {        freopen("queue25.in","r",stdin);        freopen("queue25.out","w",stdout);    }    n=read(),m=read();    for (int i=1;i<=n;i++) a[i]=read(),ha.ins(a[i],a[i]);    p[0][0]=p[0][1]=1;    for (int i=1;i<=50;i++)        p[i][0]=p[i-1][0]*A%MOD,p[i][1]=p[i-1][1]*A%Mod;    while (m--)    {        int op=read();        if (op==1)        {            int x=read(),y=read();            link(x,y);        }        if (op==2)        {            int x=read(),y=son[x];            cut(x,y);        }        if (op==3) printf("%d\n",solve());    }    return 0;}

然而有些网站跑得慢，过不了，所以就需要sam来维护。

#include <bits/stdc++.h>#define gc getchar()#define ll long long#define mod 998244353#define N 4500009#define M 10000009#define DEBUG 1#define debug() cerr<<"wrong"<<endlusing namespace std;int n,m,a[N],fa[N],Son[N],b[M];int son[N][6],fail[N],w[N];int cnt;void init(){    cnt=1;    for (int i=0;i<6;i++) son[cnt][i]=0;    fail[cnt]=w[cnt]=0;}int go(int x,int k){    if (!son[x][k]) fail[son[x][k]=++cnt]=son[fail[x]][k];    return son[x][k];}inline int read(){    int x=1;    char ch;    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;    int s=ch-'0';    while (ch=gc,ch<='9'&&ch>='0') s=s*10+ch-'0';    return s*x;}inline void link(int x,int y){    fa[y]=x,Son[x]=y;    int last=x,next=y,len1=1,len2=1,len=0;    for (int i=1;i<=48&&fa[last];i++) last=fa[last],len1++;    for (int i=1;i<=48&&Son[next];i++) next=Son[next],len2++;    for (int i=1;i<=len1+len2;i++)        b[++len]=a[last],last=Son[last];    for (int i=2,j,k;i<=50&&i<=len;i++)//i=added string's length    {        int now=1;        for (j=max(1,len1-i+2),k=i;--k;j++)            now=son[now][b[j]];        for (;j<=len&&j-i+1<=len1;j++)        {            now=go(now,b[j]);            w[now]++;            now=fail[now];        }    }}inline void cut(int x,int y){    int last=x,next=y,len1=1,len2=1,len=0;    for (int i=1;i<=48&&fa[last];i++) last=fa[last],len1++;    for (int i=1;i<=48&&Son[next];i++) next=Son[next],len2++;    for (int i=1;i<=len1+len2;i++)        b[++len]=a[last],last=Son[last];    for (int i=1,k=x;i<50&&k;k=fa[k],i++)//i=deleted string's length    {        int now=1;        for (int p=k,j=1;j<=50&&p;j++,p=Son[p])        {            now=son[now][a[p]];            if (j>i) w[now]--;//p is behind x        }    }    fa[y]=Son[x]=0;}inline int solve(){    int len=0,k;    char ch;    while (ch=gc,ch<'1'||ch>'6');    while (ch!=' ') b[++len]=ch-'1',ch=gc;    k=read();    int ans=1,now=1;    for (int i=1;i<=len;i++)    {        if (!son[now][b[i]])        {            ans=0;            break;        }        now=son[now][b[i]];        if (i>=k) ans=(ll)ans*w[now]%mod,now=fail[now];    }    return ans;}int main(){    if (DEBUG)    {        freopen("queue25.in","r",stdin);        freopen("queue25.out","w",stdout);    }    n=read(),m=read();    init();    for (int i=1;i<=n;i++)    {        a[i]=read()-1;        if (!son[1][a[i]]) fail[son[1][a[i]]=++cnt]=1;        w[son[1][a[i]]]++;    }    while (m--)    {        int op=read();        if (op==1)        {            int x=read(),y=read();            link(x,y);        }        if (op==2)        {            int x=read(),y=Son[x];            cut(x,y);        }        if (op==3) printf("%d\n",solve());    }    return 0;}

D1T3：
原问题可以转化为Sdanger≤k的答案
首先dp[i][j]表示宽度为j，前i−1行全是danger，最后sdanger≤k的概率。
所以上面的答案就是dp[1][n]
然后这个可以比较容易dp转移。
然后可以算出2到1001的dp，这是O(k2logk)的。
然后线性其次递推，用一些姿势可以算出第1行的答案。
（反正我还不会）

#include <bits/stdc++.h>#define ll long long#define mod 998244353#define N 2009using namespace std;int n,k,x,y,dp[N][N],pw[N],p,f[N],g[N],a[N],b[N],c[N];int ksm(int x,int y,int ret=1){    for (;y;y>>=1,x=(ll)x*x%mod)        if (y&1) ret=(ll)ret*x%mod;    return ret;}void mul(int n,int *a,int *b){    memset(c,0,sizeof(c));    for (int i=0;i<=n;i++)        for (int j=0;j<=n;j++)            c[i+j]=(c[i+j]+(ll)a[i]*b[j]%mod)%mod;    for (int i=2*n;i>n;i--)        for (int j=1;j<=n+1;j++)            c[i-j]=(c[i-j]+(ll)c[i]*g[j]%mod)%mod;    for (int i=0;i<=n;i++) a[i]=c[i];}int solve(int limit){    if (!limit) return ksm(mod+1-p,n);    memset(dp,0,sizeof(dp));    dp[limit+1][0]=1;    dp[limit+1][1]=pw[limit];    for (int i=limit;i;i--)    {        for (int j=0;j<=limit;j++) dp[i][j]=dp[i+1][j];        for (int j=1;(i-1)*j<=limit&&j<=limit;j++)            for (int x=1;x<=j;x++)                dp[i][j]=(dp[i][j]+(ll)dp[i+1][x-1]*dp[i][j-x]%mod*pw[i-1]%mod)%mod;    }    if (n<=limit) return dp[1][n];    for (int i=1;i<=limit+1;i++) g[i]=(ll)dp[2][i-1]*(mod+1-p)%mod;    memset(a,0,sizeof(a));    memset(b,0,sizeof(b));    a[0]=1,b[1]=1;    for (int i=n+1;i;i>>=1)    {        if (i&1) mul(limit,a,b);        mul(limit,b,b);    }//calculate the coefficients    int ans=0;    memset(f,0,sizeof(f));    f[0]=ksm(mod+1-p,mod-2);//f[0]=1/(1-p)    for (int i=1;i<=limit+1;i++)        f[i]=dp[1][i-1];    for (int i=0;i<=limit;i++) ans=(ans+(ll)f[i]*a[i]%mod)%mod;    return ans;}int main(){    scanf("%d%d%d%d",&n,&k,&x,&y);    p=(ll)x*ksm(y,mod-2)%mod;    pw[0]=mod+1-p;    for (int i=1;i<N;i++) pw[i]=(ll)pw[i-1]*p%mod;    printf("%d\n",(solve(k)-solve(k-1)+mod)%mod);    return 0;}

D2T1：
就是一道2-sat简单题。。（然而我没做出来）
枚举x为a还是b地图，然后直接2-sat模板上就好了。
时间复杂度O(2d(n+m))
（据说3d在现场也可以过）

#include <bits/stdc++.h>#define gc getchar()#define N 50009#define mid (l+r>>1)#define ll long longusing namespace std;const char ans[3][2]={{'B','C'},{'A','C'},{'A','B'}};int n,d,pos[10],num,mp[N],m,first[N<<1],number;int dfn[N<<1],low[N<<1],sta[N<<1],top,vis[N<<1],col[N<<1],col_num,cnt;int First[N<<1],Number,ds[N<<1],choose[N<<1],Ans[N];vector<int> opp[N<<1],color[N<<1];queue<int> q;struct in_edge{    int x,hx,y,hy;}Ed[N<<1];struct edge{    int from,to,next;    void add(int x,int y)    {        from=x,to=y,next=first[x],first[x]=number;    }}e[N<<2];struct Edge{    int from,to,next;    void add(int x,int y)    {        from=x,to=y,next=First[x],First[x]=Number;    }}E[N<<2];int read(){    int x=1;    char ch;    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;    int s=ch-'0';    while (ch=gc,ch<='9'&&ch>='0') s=s*10+ch-'0';    return s*x;}void get_map(){    for (int i=1;i<=(n<<1);i++) First[i]=first[i]=0;    number=Number=0;    for (int i=1;i<=m;i++)    {        int x,y;        if (mp[Ed[i].x]<2) x=(Ed[i].hx==2);        else x=(Ed[i].hx==1);        x+=2*Ed[i].x-1;        if (mp[Ed[i].y]<2) y=(Ed[i].hy==2);        else y=(Ed[i].hy==1);        y+=2*Ed[i].y-1;        if (Ed[i].x==Ed[i].y)        {            if (Ed[i].hx==Ed[i].hy) continue;            else             {                if (Ed[i].hx==mp[Ed[i].x]) continue;                else                    e[++number].add(x,4*Ed[i].x-x-1);            }        }        if (mp[Ed[i].x]==Ed[i].hx) continue;        if (mp[Ed[i].y]==Ed[i].hy)        {            e[++number].add(x,4*Ed[i].x-x-1);            continue;        }        e[++number].add(x,y);        e[++number].add(4*Ed[i].y-y-1,4*Ed[i].x-x-1);    }}void Tarjan(int x){    dfn[x]=++cnt;    low[x]=cnt;    vis[x]=1;    sta[++top]=x;    for (int i=first[x];i;i=e[i].next)    {        int temp=e[i].to;        if (!dfn[temp])        {            Tarjan(temp);            low[x]=min(low[x],low[temp]);        }        else            if (vis[temp]) low[x]=min(low[x],dfn[temp]);    }    if (dfn[x]==low[x])    {        vis[x]=0;        col[x]=++col_num;        color[col_num].push_back(x);        while (sta[top]!=x)        {            col[sta[top]]=col_num;            color[col_num].push_back(sta[top]);            vis[sta[top--]]=0;        }        top--;    }}bool solve(){    top=cnt=col_num=0;    for (int i=1;i<=(n<<1);i++)        choose[i]=ds[i]=dfn[i]=low[i]=vis[i]=0;    for (int i=1;i<=(n<<1);i++)        opp[i].clear(),color[i].clear();    for (int i=1;i<=(n<<1);i++)        if (!dfn[i]) Tarjan(i);    for (int i=1;i<=number;i++)        if (col[e[i].to]!=col[e[i].from])        {            E[++Number].add(col[e[i].to],col[e[i].from]);            ds[col[e[i].from]]++;        }    for (int i=1;i<=n;i++)    {        if (col[i*2-1]==col[i*2]) return 0;        opp[col[i*2-1]].push_back(col[i*2]);        opp[col[i*2]].push_back(col[i*2-1]);    }    while (!q.empty()) q.pop();    for (int i=1;i<=col_num;i++)        if (ds[i]==0) q.push(i);    for (int l=1;l<=col_num;l++)    {        int now=q.front();        //cout<<now<<endl;        q.pop();        if (!choose[now])        {            choose[now]=1;            for (int i=0;i<color[now].size();i++)                Ans[(color[now][i]+1)>>1]=(color[now][i]+1)&1;            for (int i=0;i<opp[now].size();i++)                choose[opp[now][i]]=2;        }        for (int i=First[now];i;i=E[i].next)            if (!(--ds[E[i].to])) q.push(E[i].to);    }    return 1;}void print(){    for (int i=1;i<=n;i++) putchar(ans[mp[i]][Ans[i]]);    puts("");}int main(){    n=read(),d=read();    for (int i=1;i<=n;i++)    {        char ch;        while (ch=gc,ch!='x'&&ch!='a'&&ch!='b'&&ch!='c');        if (ch=='x') pos[++num]=i;        else mp[i]=ch-'a';    }    m=read();    for (int i=1;i<=m;i++)    {        char ch;        Ed[i].x=read();        while (ch=gc,ch!='A'&&ch!='B'&&ch!='C');        Ed[i].hx=ch-'A';        Ed[i].y=read();        while (ch=gc,ch!='A'&&ch!='B'&&ch!='C');        Ed[i].hy=ch-'A';    }    for (int now=0;now<(1<<d);now++)    {        for (int i=1;i<=d;i++) mp[pos[i]]=(now>>(i-1))&1;        get_map();        if (solve())        {            print();            return 0;        }    }    puts("-1");    return 0;}

D2T2&&D2T3：
一道费用流神题，一道计算几何神题，反正我不会写。。。（T2可能还有一丝希望）
（未完待续）